Question

If H is a subgroup of a group G, let X designate the set of all the left cosets of H in G.

For each element a ∈G, define pₐ :X→X as follows:
pₐ (xH)=(ax)H
1. Prove that each pₐ is a permutation of X.

Answer 1

To prove that p₀ is a permutation of X, we show that it is injective by demonstrating that p₀(xH) = p₀(yH) implies x = y, and we prove surjectivity by finding a pre-image for every coset bH in X. Therefore, p₀ is bijective and consequently a permutation.

Step-by-step explanation:

To prove that p₀ is a permutation of X, we need to show that it is a bijective function, meaning it is both injective (one-to-one) and surjective (onto). We consider X to be the set of all left cosets of H in G, where G is a group and H is a subgroup.

• Injective: Assume p₀(xH) = p₀(yH). This means (ax)H = (ay)H, and thus a-1(ax)H = a-1(ay)H, which simplifies to xH = yH. Therefore, x and y are in the same coset, and x=y, proving injectivity.
• Surjective: For any coset bH in X, we can find a coset a-1bH such that p₀(a-1bH) = (a(a-1b))H = bH, which shows that every element in X has a pre-image under p₀, proving surjectivity.

Since p₀ is both injective and surjective, it is indeed a permutation of X.